1 Description #
source: https://leetcode.com/problems/missing-number/
Given an array nums
containing n distinct numbers in the range [0, n]
, return the only number in the range that is missing from the array.
Example 1:
Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
Example 2:
Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
Example 3:
Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
Constraints:
n == nums.length
1 <= n <= 10^4
0 <= nums[i] <= n
- All the numbers of
nums
are unique.
Follow up: Could you implement a solution using only O(1)
extra space complexity and O(n)
runtime complexity?
2 Solution #
#include <vector>
class Solution {
public:
int missingNumber(vector<int>& nums) {
// Time complexity: O(n), n is the size of nums.
// Space complexity: O(1)
int n = nums.size();
int expected_sum = n * (n + 1) / 2.0;
for(const auto& num: nums){
expected_sum -= num;
}
return expected_sum;
}
};